Problem 1. (§3.2, #18) The plane x 3y z = 12 is parallel to the plane x 3y z = 0 in Problem 17. One particular point on this plane is (12, 0, 0). All points on the plane have the form (fill in the first components)
Solution. The equation x = 12 + 3y + z says it all:
Problem 2. (§3.2, #24) (If possible. . . ) Construct a matrix whose column space contains (1, 1, 0) and (0, 1, 1) and whose nullspace contains (1, 0, 1) and (0, 0, 1).
Solution. Not possible : Such a matrix A must be 3 × 3. Since the nullspace is supposed to contain two independent vectors, A can have at most 3 2 = 1 pivots. Since the column space is supposed to contain two independent vectors, A must have at least 2 pivots. T hese conditions cannot both be met!
Problem 3. (§3.2, #36) How is the nullspace N( C ) related to the spaces N( A ) and N( B ), if C = A ? B
Solution. N( C ) = N( A ) ∩ N( B ) just the intersection: Indeed, A x B x C x = so that C x = 0 if and only if A x = 0 and B x = 0. (...and as a nitpick, it wouldn’t be quite sloppy instead write “if and only if A x = B x = 0”—those are zero vectors of potentially different length, hardly equal). Q
Problem 4. (§3.2, #37) Kirchoff’s Law says that current in = current out at every node. This network has six currents y1 , . . . , y6 (the arrows show the positive direction, each yi could be positive or negative). Find the four equations A y = 0 for Kirchoff’s Law at the four nodes. Find three special solutions in the nullspace of A
Solution. The four equations are, in order by node,
Adding the last three rows to the first eliminates it, and shows that we have three “pivot variables”
and three “free variables”
special solutions by back-substitution
Problem 5. (§3.3, #19) Suppose A and B are n by n matrices, and AB = I . Prove from rank( AB ) ≤ rank(A) that the rank of A is n So A is invertible and B must be its two-sided inverse (Section 2.5). Therefore BA = I (which is not so obvious!).
Solution. Since A is n by n, rank(A) ≤ n and conversely
n = rank(I n ) = rank( AB ) ≤ rank(A)
The rest of the problem statement seems to be “commentary,” and not further things to do. Q
Problem 6. (§3.3, #25) Neat fact Every m by n matrix of rank r reduces to ( m by r ) times ( r by n ):
A = (pivot columns of A ) (first r rows of R )) = (C O L ) (R OW )
Write the 3 by 4 matrix A in equation (1) at the start of this section as the product of the 3 by 2 matrix from the pivot columns and the 2 by 4 matrix from R
Problem 7. (§3.3, #27) Suppose R is m by n of rank r, with pivot columns first:
(a) What are the shapes of those four blocks?
(b) Find a right-inverse B with RB = I if r = m
(c) Find a left-inverse C with CR = I if r = n
(d) What is the reduced row echelon form of R T (with shapes)?
(e) What is the reduced row echelon form of R T R (with shapes)?
Prove that R T R has the same nullspace as R Later we show that A T A always has the same nullspace as A (a valuable fact).
(b) In this case
(c) In this case
(d) Note that
(e) Note that
Performing row operations doesn’t change the nullspace, so that N( A ) = N( rref ( A )) for any matrix A . So, N( A ) = N( R T R ) by (e). Q
Problem 8. (§3.3, #28) Suppose you allow elementary column operations on A as well as elementary row operations (which get to R ). What is the “row-and-column reduced form” for an m by n matrix of rank r?
Solution. After getting to R we can use the column operations to get rid of F , and get to
Problem 9. (§3.3, #17 – Optional)
(a) Suppose column j of B is a combination of previous columns of B Show that column j of AB is the same combination of previous columnd of AB Then AB cannot have new pivot columns, so rank( AB ) ≤ rank(B).
(b) Find A 1 and A 2 so that rank(
Solution. (a) That column j of B is a combination of previous columns of B means precisely that there exist numbers a1 , , aj 1 so that each row vector x = ( x i ) of B satisfies the linear relation
The rows of the matrix AB are all linear combinations of the rows of B , and so also satisfy this linear relation. So, column j is the same combination of previous columns of AB , as desired. Since a column is pivot column precisely when it is not a combination of previous columns, this shows that AB cannot have previous columns and the rank inequality.
(b) Take A 1 = I 2 and A 2 = 02 (or for a less trivial example A 2 = 1 1 ). 1 1 Q
Problem 10. (§3.4, #13) Explain why these are all false:
(a) The complete solution is any linear combination of x p and x n
(b) A system A x = b has at most one particular solution.
(c) The solution x p with all free variables zero is the shortest solution (minimum length ǁxǁ). Find a 2 by 2 counterexample.
(d) If A is invertible there is no solution x n in the nullspace.
Solution. The coefficient of x p must be one.
(a) If x n ∈ N( A ) is in the nullspace of A and x p is one particular solution, then x p + x n is also a particular solution.
(c) Lots of counterexamples are possible. Let’s talk about the 2 by 2 case geometrically: If A is a 2 by 2 matrix of rank 1, then the solutions to A x = b form a line parallel to the line that is the nullspace. We’re asking that this line’s closest point to the origin be somewhere not along an axis. The line x + y = 1 gives such an example.
Explicitly, let
Then, ǁ x p ǁ = 1/√ 2 < 1 while the particular solutions having some coordinate equal to zero are (1, 0) and (0, 1) and they both have ǁ · ǁ = 1.
(d) There’s always x n = 0.
Problem 11. (§3.4, #25) Write down all known relations between r and m and n if A x = b has
(a) no solution for some b
(b) infinitely many solutions for every b
(c) exactly one solution for some b, no solution for other b
(d) exactly one solution for every b.
Solution.The system has less than full row rank: r < m
(a) The system has full row rank, and less than full column rank: m = r < n
(b) The system has full column rank, and less than full row rank: n = r < m
(c) The system has full row and column rank (i.e., is invertible): n = r = m
Problem 12. (§3.4, #28) Apply Gauss-Jordan elimination to U x = 0 and U x = c. Reach R x = 0 and R x =
Solution. Let me just say to whoever’s reading: The problem statement is confusing as written!! In any case, I think the desired response is:
so that
Problem 13. (§3.4, #35) Suppose K is the 9 by 9 second difference matrix (2’s on the diagonal, -1’s on the diagonal above and also below). Solve the equation K x = b = (10, . . . , 10). If you graph x 1 , , x 9 above the points 1, , 9 on the x axis, I think the nine points fall on a parabola.
Solution. Here’s some MATLAB code that should do this:
K = 2*eye(9) + diag(-1*ones(1,8),1) + b = 10*ones(9,1); x = K \ b
It gives back that
And for fun, the graph is indeed parabola-like:
diag(-1*ones(1,8),-1);
Q
Problem 14. (§3.4, #36) Suppose A x = b and C x = b have the same (complete) solutions for every b. Is it true that A = C ?
Solution. Yes In order to check that A = C as matrices, it’s enough to check that A y = C y for all vectors y of the correct size (or just for the standard basis vectors, since multiplication by them “picks out the columns”). So let y be any vector of the correct size, and set b = A y. Then y is certainly a solution to A x = b, and so by our hypothesis must also be a solution to C x = b; in other words, C y = b = A y. Q
